Considering this question, I was struck by the idea that:
For a graph $G$ that is a tree, the number of degree-$1$ nodes exceeds the number of nodes of degree $3$ or higher.
.. which would fairly directly solve that question.
The intuition is as follows: each degree-$3$ node adds a branch to the tree, which must also add a degree-$1$ node. Higher degree nodes add branches per extra degree. This intuition adds a numerical prediction: $$ \sum_{k\in H}(d(k)-2) = |L|-2$$ where $H$ is the set of vertices in $G$ with degree $3$ or higher and $L$ is the set of degree-$1$ nodes, with two leaf nodes being allocated to a simple unbranched tree.
Can anyone produce or reference a more formal proof?